我试图通过编写一个在输入缓冲区中找到值的平方的内核来学习OpenCL中的clEnqueueMapBuffer，但是只使用clEnqueueMapBuffer在输出缓冲区中一次返回两个项目。据我了解，此函数返回主机内存中的指针，指针指向设备中的缓冲区内存。然后clEnqueueUnmapMemObject必须取消映射此缓冲区以允许内核继续其计算。现在，当我调用clEnqueueMapBuffer时，它返回随机数据。
这是我的内核
__kernel void testStream(
__global int *input_vector,
__global int *output_vector,
__global int *mem_flag) // informs the host when the workload is finished
{
mem_flag[0] = 1;
}
和我的来源
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <CL/opencl.h>
#include "utils.h"
int main(void)
{
unsigned int n = 24;
int BUFF_SIZE = 2;
// Input and output vectors
int num_bytes = sizeof(int) * n;
int *output_buffer = (int *) malloc(num_bytes);
int output_buffer_offset = 0;
int *mapped_data = NULL;
// use mapped_flag for determining if the job on the device is finished
int *mapped_flag = NULL;
int *host_in = (int *) malloc(num_bytes);
int *host_out = (int *) malloc(num_bytes);
// Declare cl variables
cl_mem device_in;
cl_mem device_out;
cl_mem device_out_flag;
// Declare cl boilerplate
cl_platform_id platform = NULL;
cl_device_id device = NULL;
cl_command_queue queue = NULL;
cl_context context = NULL;
cl_program program = NULL;
cl_kernel kernel = NULL;
// Located in utils.c -- the source is irrelevant here
char *kernel_source = read_kernel("kernels/test.cl");
// Initialize host_in
int i;
for (i = 0; i < n; i++) {
host_in[i] = i + 1;
}
// Set up opencl
cl_int error;
error = clGetPlatformIDs(1, &platform, NULL);
printf("clGetPlatformIDs: %d\n", (int) error);
error = clGetDeviceIDs(platform, CL_DEVICE_TYPE_CPU, 1, &device, NULL);
printf("clGetDeviceIDs: %d\n", (int) error);
context = clCreateContext(NULL, 1, &device, NULL, NULL, &error);
printf("clCreateContext: %d\n", (int) error);
queue = clCreateCommandQueue(context, device, 0, &error);
printf("clCreateCommandQueue: %d\n", (int) error);
program = clCreateProgramWithSource(context, 1,
(const char**)&kernel_source, NULL, &error);
printf("clCreateProgramWithSource: %d\n", (int) error);
clBuildProgram(program, 0, NULL, NULL, NULL, NULL);
kernel = clCreateKernel(program, "testStream", &error);
printf("clCreateKernel: %d\n", (int) error);
// Create the buffers
device_in = clCreateBuffer(context, CL_MEM_READ_ONLY,
num_bytes, NULL, NULL);
device_out = clCreateBuffer(context,
CL_MEM_WRITE_ONLY,
sizeof(int) * BUFF_SIZE, NULL, NULL);
device_out_flag = clCreateBuffer(context,
CL_MEM_WRITE_ONLY,
sizeof(int) * 2, NULL, NULL);
// Write the input buffer
clEnqueueWriteBuffer(
queue, device_in, CL_FALSE, 0, num_bytes,
host_in, 0, NULL, NULL);
// Set the kernel arguments
error = clSetKernelArg(kernel, 0, sizeof(cl_mem), &device_in);
error = clSetKernelArg(kernel, 1, sizeof(cl_mem), &device_out);
error = clSetKernelArg(kernel, 2, sizeof(cl_mem), &device_out_flag);
// Execute the kernel over the entire range of data
clEnqueueNDRangeKernel(queue, kernel, 1, NULL,
(const size_t *) &n, NULL, 0, NULL, NULL);
// Map and unmap until the flag is set to true
int break_flag = 0;
while(1) {
// Map the buffers
mapped_data = (int *) clEnqueueMapBuffer(
queue, device_out, CL_TRUE, CL_MAP_READ, 0,
sizeof(int) * BUFF_SIZE, 0, NULL, NULL, &error);
mapped_flag = (int *) clEnqueueMapBuffer(
queue, device_out_flag, CL_TRUE, CL_MAP_READ, 0,
sizeof(int) , 0, NULL,NULL, &error);
// Extract the data out of the buffer
printf("mapped_flag[0] = %d\n", mapped_flag[0]);
// Set the break_flag
break_flag = mapped_flag[0];
// Unmap the buffers
error = clEnqueueUnmapMemObject(queue, device_out, mapped_data, 0,
NULL, NULL);
error = clEnqueueUnmapMemObject(queue, device_out_flag, mapped_flag,
0, NULL, NULL);
if (break_flag == 1) {break;}
usleep(1000*1000);
}
return 0;
}
当我运行程序时，我得到的输出类似于
clGetPlatformIDs: 0
clGetDeviceIDs: 0
clCreateContext: 0
clCreateCommandQueue: 0
clCreateProgramWithSource: 0
clCreateKernel: 0
mapped_flag[0] = 45366144
mapped_flag[0] = 45366144
mapped_flag[0] = 45366144
mapped_flag[0] = 45366144
mapped_flag[0] = 45366144
为什么会这样？
我在内核3.13.7-100.fc19.x86_64上使用带有fedora 19 64位的HP dm1z运行此代码。这是clinfo的输出
Number of platforms: 1
Platform Profile: FULL_PROFILE
Platform Version: OpenCL 1.2 AMD-APP (1214.3)
Platform Name: AMD Accelerated Parallel Processing
Platform Vendor: Advanced Micro Devices, Inc.
Platform Extensions: cl_khr_icd cl_amd_event_callback cl_amd_offline_devices
Platform Name: AMD Accelerated Parallel Processing
Number of devices: 2
Device Type: CL_DEVICE_TYPE_GPU
Device ID: 4098
Board name: AMD Radeon HD 6310 Graphics
Device Topology: PCI[ B#0, D#1, F#0 ]
Max compute units: 2
Max work items dimensions: 3
Max work items[0]: 256
Max work items[1]: 256
Max work items[2]: 256
Max work group size: 256
Preferred vector width char: 16
Preferred vector width short: 8
Preferred vector width int: 4
Preferred vector width long: 2
Preferred vector width float: 4
Preferred vector width double: 0
Native vector width char: 16
Native vector width short: 8
Native vector width int: 4
Native vector width long: 2
Native vector width float: 4
Native vector width double: 0
Max clock frequency: 492Mhz
Address bits: 32
Max memory allocation: 134217728
Image support: Yes
Max number of images read arguments: 128
Max number of images write arguments: 8
Max image 2D width: 16384
Max image 2D height: 16384
Max image 3D width: 2048
Max image 3D height: 2048
Max image 3D depth: 2048
Max samplers within kernel: 16
Max size of kernel argument: 1024
Alignment (bits) of base address: 2048
Minimum alignment (bytes) for any datatype: 128
Single precision floating point capability
Denorms: No
Quiet NaNs: Yes
Round to nearest even: Yes
Round to zero: Yes
Round to +ve and infinity: Yes
IEEE754-2008 fused multiply-add: Yes
Cache type: None
Cache line size: 0
Cache size: 0
Global memory size: 201326592
Constant buffer size: 65536
Max number of constant args: 8
Local memory type: Scratchpad
Local memory size: 32768
Kernel Preferred work group size multiple: 32
Error correction support: 0
Unified memory for Host and Device: 1
Profiling timer resolution: 1
Device endianess: Little
Available: Yes
Compiler available: Yes
Execution capabilities:
Execute OpenCL kernels: Yes
Execute native function: No
Queue properties:
Out-of-Order: No
Profiling : Yes
Platform ID: 0x00007fd434852fc0
Name: Loveland
Vendor: Advanced Micro Devices, Inc.
Device OpenCL C version: OpenCL C 1.2
Driver version: 1214.3
Profile: FULL_PROFILE
Version: OpenCL 1.2 AMD-APP (1214.3)
Extensions: cl_khr_global_int32_base_atomics cl_khr_global_int32_extended_atomics cl_khr_local_int32_base_atomics cl_khr_local_int32_extended_atomics cl_khr_3d_image_writes cl_khr_byte_addressable_store cl_khr_gl_sharing cl_ext_atomic_counters_32 cl_amd_device_attribute_query cl_amd_vec3 cl_amd_printf cl_amd_media_ops cl_amd_media_ops2 cl_amd_popcnt cl_amd_image2d_from_buffer_read_only
另外，值得注意的是，当我开始玩OpenCL时，我运行了一个测试程序来计算内部产品，但这给出了奇怪的结果。最初我虽然这是程序错误而忘记了它，但OpenCL实现是否有可能出错？如果它有帮助，OpenGL实现会有多个错误，导致随机数据块出现在我的桌面背景上，但这也可能是Linux问题。
答案 0 :(得分：4)
您将NULL作为全球工作规模传递给clEnqueueNDRangeKernel来电：
// Execute the kernel over the entire range of data
clEnqueueNDRangeKernel(queue, kernel, 1, NULL, NULL, NULL, 0, NULL, NULL);
如果您正在检查此调用返回的错误代码（总是应该），则会收到与CL_INVALID_GLOBAL_WORK_SIZE对应的错误代码。您始终需要指定全局工作大小，因此您的调用应如下所示：
// Execute the kernel over the entire range of data
size_t global[1] = {1};
error = clEnqueueNDRangeKernel(queue, kernel, 1, NULL, global, NULL, 0, NULL, NULL);
// check error == CL_SUCCESS!
您对map和unmap缓冲区的调用很好;我已使用上述修复程序对此代码进行了测试，它对我有用。
修复上述问题的更新代码如下所示：
unsigned int n = 24;
...
// Execute the kernel over the entire range of data
clEnqueueNDRangeKernel(queue, kernel, 1, NULL,
(const size_t *) &n, NULL, 0, NULL, NULL);
这不是将全局工作大小参数传递给内核的安全方法。例如，unsigned int n变量可能占用32位，而size_t可能是64位。这意味着当您传递n的地址并转换为const size_t*时，实现将读取64位值，该值将包含32位n加上32位其他位有一些任意的价值。您应该将n分配给size_t变量，然后再将其传递给clEnqueueNDRangeKernel，或者将其更改为size_t本身。
这可能与您遇到的问题有关，也可能与之无关。例如，您可能无意中启动了巨大的数量的工作项，这可能解释了为什么代码似乎在CPU上阻塞。
答案 1 :(得分：1)
以下是我的一些评论：
clEnqueueMapBuffer将会中断内核的执行。我不认为这是正确的。一旦命令启动执行，它就会一直运行直到它完成（或失败......）。可以同时启动多个命令，但是在内核仍处理它时尝试读取某些数据将导致未定义的行为。此外，创建命令队列的方式不允许您同时运行多个命令。创建队列时需要使用CL_QUEUE_OUT_OF_ORDER_EXEC_MODE_ENABLE属性才能允许。mem_flag[0] = 1;因为所有的工作项都会写到同一个地方。 （我猜你只发布了你内核的一部分。检查你是否有其他类似的声明......如果你发布整个内核代码，那么它实际上很有用。）